Question

Please help!!!

Answer 1

Is the question asking for all values of k for which f(x) exists for all x? Is that what is meant by continuously defined?

@nguyen8ki

Answer 2

yes

Answer 3

read the question wrong, sec.

Answer 4

I thought your answer was number e lol

Answer 5

lololol

Answer 6

It should be (0, 2].

Answer 7

show me how you got the answet, please

Answer 8

Wait sorry, I mean [-2, 0)

Answer 9

typed it backwards lol

Answer 10

K so what you're trying to do here is finding a value of k such that the quadratic under the square root is always non-negative and the denominator is always non-zero.

Let's first consider the case of the denominator being always non-zero. This happens whenever k is negative, i.e k is in the interval (-infinity, 0). Whe k is negative, x^2 - k = x^2 - (-k) = x^2 + k. What this will mean is that the denominator will now always be positive if k is in (-infinity, 0).

Now we need to consider the case such that the quadratic under the square is always non-negative. This is required because we cannot take the square root of a negative number which means the value of k should be such that the quadratic under the square root is positive for all x. This will happen if the value of k is in the interval (infinity, 0) but it cannot be lower than -2. If it was then there will be some values of x that for which the quadratic will yield negative values and we cannot take the square root of negative values. Only values of k in [-2, 0) produce a quadratic that is non-negative for all x and so that's the answer.

Answer 11

I see what you're saying. thank you so much.

Answer 12

Hang on.
Let's first consider the case of the denominator being always non-zero. This happens whenever k is negative, i.e k is in the interval (-infinity, 0). Whe k is negative, x^2 - k = x^2 - (-k) = x^2 + k. What this will mean is that the denominator will now always be positive if k is in (-infinity, 0).
I think your logic was a little uncertain here. What if I choose k=1 and x=2? The denominator is still positive.